I am psyched you made this video! I use this theorem in my videos all the time but haven't really wanted to prove it!
@Bayerwaldler4 жыл бұрын
I like your videos but I noticed that you very often say "by the dominated convergence theorem..." I bet it is trivial most of the time, but it would be nice if you could show the dominating function from time to time.
@MichaelPennMath4 жыл бұрын
@@Bayerwaldler I totally agree, I am for sure taking advantage a little bit... I'll make a deal, next time I use this theorem I will show the dominating function.
@jkid11344 жыл бұрын
I'm sure there are some interesting examples and then also some trivial examples (like sine) that would both be worth the time to mention the dominating function. I say you probably want to prove the theorem itself once, so any questions can be met with a video link. Same with Fubini and anything you want to do with exchanging the order of two sums or integrals. There is NOTHING worse as a math student than hearing "this works and I won't show you why" Edit to be a little less coy: I am a big fan, and I would love a video about what happens to the bounds when sums and integrals are exchanged, and why it is okay to do.
@TheMauror224 жыл бұрын
You should both prove it :( yours and Dr. peyams's are my favorite youtube channels
@moshadj4 жыл бұрын
its still not proved. please do!
@blackpenredpen4 жыл бұрын
Hello, Analysis! I can’t believe that we meet again since 2012. Hope you are doing well. I remember what you did to me but now there’s Dr. P you won’t be able to do the same again. Ok, bye.
@drpeyam4 жыл бұрын
Hahahaha
@kingfrozen42576 ай бұрын
after studying 10years of pure math, I'm 100% sure DCT is my fav theorem! the assumptions are so mild that you can't pass on loving it!
@TheTKPizza4 жыл бұрын
Thanks for the video man, I just finished my ODE lecture and you starting with a bunch of the basics for PDEs right now is just a blessing. You are a great teacher.
@Try-tu4zq4 жыл бұрын
More videos of this kind please, Dr. PEyam.
@XAE-yc9rr4 жыл бұрын
8:40 I chuckled at the Dominican Central Time part, and I'm half Dominican! Great analysis video.
@fractalfan4 жыл бұрын
12:50 A more rigorous explanation of why we can replace n goes to infinity with h goes to 0 is that f(h) converges to y as h goes to 0 is equivalent to f (h_n) goes to y for any sequence (h_n) where h_n goes to 0 as n goes to infinity.
@laurensiusfabianussteven65184 жыл бұрын
this whole time, finally a worthy explanation
@tomatrix75254 жыл бұрын
Definitely the most unknowingly used theorem I’ve come across.
@J_Stockhausen3 жыл бұрын
Grande Dr peyam. Lo más grande de todo el mundo.
@ScholarStream_254 жыл бұрын
Smooth and brilliant
@jonathangrey63544 жыл бұрын
finally someone made a video about this
@Brien8314 жыл бұрын
This comes in really handy, writing my german Analysis II exam on friday.
@drpeyam4 жыл бұрын
Viel Glück!!!
@thomasjefferson6225 Жыл бұрын
okay this makes sense now i understand whats going on with the PDE course
@princeardalan3 жыл бұрын
A fantastic video! Thanks Peyam!
@jessehammer1234 жыл бұрын
This is way better than my video’s coverage of the DCT! Not that I’m complaining. It’s really, really good.
@WoWSchockadin4 жыл бұрын
Using the Dirac Delta Distribution as a countexample for a property of a function is a bit evil! :-P
@LeilaRmaths2 жыл бұрын
greate video with very clear explanation, thank you very much
@ArtutMTMartins Жыл бұрын
Yass! the video I was looking for
@drpeyam Жыл бұрын
😄
@minwooyoo11672 жыл бұрын
Thank you so much.
@anthonyymm5113 жыл бұрын
The fundamental theorem of analysis!
@dgrandlapinblanc2 жыл бұрын
Good. Thank you very much.
@yilmazkaraman2563 жыл бұрын
Good explanation
@sarvagyagupta17443 жыл бұрын
Thank you for sharing this video and the explanation is amazing. This somewhat reminds me of Line integral and how under another function, we find the integral. Is there a relationship between the two?
@thomasborgsmidt98014 жыл бұрын
I have to see this video a couple of times.
@hewaselman67742 жыл бұрын
well done
@martinepstein98264 жыл бұрын
My favorite sequence of functions where you can't interchange the limit and integral is f_n(x) = 1 for x in [n, n+1) and 0 otherwise. I think I heard it called the "boxcar" sequence. It's like you're chasing after a runaway boxcar of area 1 to take its integral but it keeps rolling away.
@drpeyam4 жыл бұрын
I like that!!!
@benjaminbrat39224 жыл бұрын
Thank you very much! Brings back memories of one of my most fun maths class. By the way, that's the first time I see you use computer whiteboard, what is your setup for that? Onenote with mouse and keyboard :p ?
@benjaminbrat39224 жыл бұрын
Subsequent question: you seem very chill about the hypothesis verification of the DCT. I don't think that's carelessness, which means that the DCT covers most cases in a nwarly obvious way. Is there a set of functions that are very useful in order to bound function limits? Could you apply the DCT to the pathological case of point wise function you give as an example N*Ind(0,1/N), or give more context? I feel like I am missing a puzzle piece
@drpeyam4 жыл бұрын
I’m using the microsoft whiteboard + zoom
@drpeyam4 жыл бұрын
And here we cannot bound n 1(0,1/n) because otherwise we could pass in the limit!
@iabervon4 жыл бұрын
This sounds like it would be useful for proving properties of the discrete cosine transform, but it would be really confusing namewise.
@di-dah-ditcharlie65114 жыл бұрын
Question: Why do we assume the f_n converges to 0 point wise, as by the definition of the indicator function as n -> inf f_n (x) = 0 (expect zero), for zero it seems (intuitively) [n -> inf ] n -> f_n(0) = inf. This seems to be like a delta function (For which the integral would be 1?), it will be really helpful if you can explain / point towards an explanation of what i am missing. Thanks a lot for this awesome videos.
@drpeyam4 жыл бұрын
It’s because we’re using the open internal (0,1/n). If we used the closed interval [0,1/n] then I’d agree with you
@di-dah-ditcharlie65114 жыл бұрын
@@drpeyam Thanks, I get it now.
@jksmusicstudio14393 жыл бұрын
@@di-dah-ditcharlie6511 even in the case you mentioned the integral of the limit function is still 0, because it might be infinity at a point, but that only gives us a vertical line under the graph of f, which still has area 0. Is like integrating 0 from 0 to infinity, the region under the graph here is the x-axis, which ofc has area 0. That being said, that's how riemann and lebesgue integrals work. There are other interpretations which may give integral = 1 as you expected (check out distributions for more details on that)
@salauddinshakil3403 Жыл бұрын
Is it a complete proof?
@adamlimani15284 жыл бұрын
When you are talking about integral, I assume you mean in the sense of Riemann. If that is the case, then you need to add the additional assumption that the pointwise limiting function f is Riemann-integrable as well.
@drpeyam4 жыл бұрын
No, Lebesgue
@adamlimani15284 жыл бұрын
@@drpeyam Ah, okay! In that case, good content!
@moshadj4 жыл бұрын
So in Analysis 2 we studied passing the limit under the integral sign with uniform convergence. Does the Dominated Convergence Theorem apply to more general functions?
@drpeyam4 жыл бұрын
This is more general. If fn converges uniformly to f, then it is bounded by an integrable function g (I think)
@thedoublehelix56614 жыл бұрын
finally!
@paulestrada9614 жыл бұрын
Is the absolute value around f_N(x) necessary when we are comparing it to some other function g(x) for each N within the statement of the Dominated Convergence Theorem?
@drpeyam4 жыл бұрын
Yes because fn could be negative. Think for instance -n 1(0,1/n). Then it is bounded above by 0 but the DCT doesn’t hold there
@paulestrada9614 жыл бұрын
@@drpeyam Thank you. I may have answered my own question initially after posting this comment with what I wrote and appreciate the example to help clarify and reinforce my original intuition. I look forward to more videos :).
@tgx35294 жыл бұрын
DCT theorem is only for Lebesgue integrals?Is it Lebesgue theorem? And if I want to apply Lebesgue ( Dominated convergence Theorem)sentence in combination with MVT, I need to make a measurable space was interval and functions on this interval continuous, is MVT Lagrange ?
@drpeyam4 жыл бұрын
Yes, Lebesgue integrals. And yes, you’d need it to be differentiable etc
@tgx35294 жыл бұрын
@@drpeyam Yes, thank you.
@ashuthoshbharadwaj67034 жыл бұрын
Wouldnt the example function converge to a dirac delta?? It had a constant area of 1 for all N
@Bayerwaldler4 жыл бұрын
But Dirac delta is not a function, it is a so called distribution. I know: Physicists used to (or maybe they still do?) call it the "Delta function" but mathematically speaking it is not a function.
@ashuthoshbharadwaj67034 жыл бұрын
@@Bayerwaldler Yeah that makes sense I guess, but I'm still not convinced with calling it a zero everywhere, it has to be infinite at x = 0
@pierreabbat61574 жыл бұрын
The functions are all 0 at 0. If they were n on [0,1/n) instead of (0,1/n), they would converge to the Dirac delta. But then they wouldn't converge pointwise, since f[n](0) goes to ∞.
@ashuthoshbharadwaj67034 жыл бұрын
@@pierreabbat6157 Ahhh thats it! Thank you so much for clearing that up!
@jksmusicstudio14393 жыл бұрын
@@ashuthoshbharadwaj6703 Even in the case Pierre mentioned, the integral of the limit is still 0 as long as you view the limit as a function which can take infinity as a value too (measure theory allows for suchs functions). The idea is that althought f is infinity at 0, the "width of the corresponding rectangle" (or rigorously the measure of the set which contains only 0) is 0, thus the region that is under f is just a vertical line at 0, which ofc has area 0. Is like integrating 0 over all the reals, here the width is infinite but the height is 0 and as we all know the integral is 0. Now if you go distributions'-land, that's another story...
@GhostyOcean4 жыл бұрын
So since you didn't mention the domain in the theorem, I assume that means that the domain of g is the same as f and is a subset of R?
@drpeyam4 жыл бұрын
Yeah
@jksmusicstudio14393 жыл бұрын
Technically the domain could be any measure space, which is one of the many reasons Lebesgue integral is so powerful (another reason is the DCT itself). Also it is enough to suppose that the domain of f is a subset of the domain of g (which is kinda obvious if you draw the graphs).
@leickrobinson51864 жыл бұрын
Hey, Dr. P. Nice video, but one nitpick. I would avoid reusing the same letters/identifiers in different and contradictory ways (as you do here with f, g, and c). This would avoid introducing unnecessary confusion (particularly for those who are learning this for the first time). There are a lot of other letters to choose from instead! Cheers! :-D
@davidkwon18724 жыл бұрын
9:40 Why don’t we just say, lim l f_n(x) l < infinity? Rather than using g(x). What would be a counter example when lim f_n is finite, lim Integral f_n =/ integral lim f ?
@drpeyam4 жыл бұрын
Well you could just define g(x) as lim |fn(x)| so it’s equivalent, but it’s important that g be integrable
@davidkwon18724 жыл бұрын
Dr Peyam Thank you professor!
@thomasborgsmidt98014 жыл бұрын
Hmm..... lets say we have a function: g(x) = (x^5) + (x^4) + (x^2) and a function f(x) = (x^5) + (x^3) + x then for x going to + infinity the difference g(x)-f(x) = x^5 - x^5 + x^4 - x^3 + x^2 - x = 0 + x^3 (x-1) + x (x-1) = (x-1)*x* (x^2 -1) for very large positive x..... Perhaps the ratio would be more informative? f(x)/g(x) = [x * (x^4 + x^2 + 1)] / [ (x^2) * (x^3 + x^2 + 1)] ?
@abdellatifdz8748 Жыл бұрын
👌 thx
@uroskosmac12654 жыл бұрын
Is this theorem eqivavalet to uniform convergence of f_n, and if not does it imply uniform convergence
@fractalfan4 жыл бұрын
If the limit function f is bounded and we are considering functions on a set with finite measure, then uniform convergence implies dominated convergence for n sufficiently large, since then we have f(x)-e
@uroskosmac12654 жыл бұрын
Aleksandra Bozovic thanks that was helpful
@matekichba66403 жыл бұрын
16:04 Mi español interior ha despertado 😆
@Happy_Abe4 жыл бұрын
Why isn't the limit of the function a delta function?
@martinepstein98264 жыл бұрын
The delta "function" isn't a function. It's something called a distribution.
@Happy_Abe4 жыл бұрын
@@martinepstein9826 I still have my question then, it should converge to that distribution
@martinepstein98264 жыл бұрын
@@Happy_Abe It doesn't though. Forget about what the sequence looks like graphically and focus on the definition of pointwise convergence; the limit of the sequence [f_1(0), f_2(0), f_3(0), ...] is 0. Therefore f(0) = 0 and f is not the Dirac delta distribution.
@Happy_Abe4 жыл бұрын
@@martinepstein9826 I was focusing on the graphical element yeah, thanks.
@martinepstein98264 жыл бұрын
8:36 Why is there a theorem about Dominican Central Time? 8:39 Oh nvmd